Solution:

Concept: Kinematics equations involving differentiation.
Formula: Velocity \(v = \frac{ds}{dt}\) and acceleration \(a = \frac{dv}{dt}\).
Solution: Given \(s = t^3 - 6t^2 + 3t + 4\). Then \(v = 3t^2 - 12t + 3\) and \(a = 6t - 12\). Setting \(a=0\) gives \(t=2\text{ s}\). Substituting \(t=2\text{ s}\) into \(v\) gives \(v = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9\text{ m/s}\).